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rosijanka [135]
3 years ago
13

An object is moving at 10 m/s slows to 6 m/s in 4 seconds. What is its average acceleration?? Help pls!!

Physics
1 answer:
Ivahew [28]3 years ago
6 0
You need to find the acceleration . a=v-vo/t
a=6-10/4
a=-4/4
a=-1m/s/s or -1m/s^2
You’re welcome :)
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What kind of waves are present during an earthquake? ...
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Answer:

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3 years ago
A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio
Step2247 [10]

The final volume of the gas is 144.25 L

Explanation:

For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by

W=p\Delta V = p(V_f - V_i)

where

p is the pressure of the gas

V_i is the initial volume

V_f is the final volume

For the gas in the cylinder in this problem,

p = 2.00 atm

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And we also know the work done,

W = 288 J

So we can solve the equation for V_f, the final volume:

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Learn more about ideal gases:

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brainly.com/question/7316997

brainly.com/question/3658563

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7 0
3 years ago
Find the area under the standard normal curve to the right of z=0.49. round your answer to four decimal places, if necessary.
VikaD [51]

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

<h3>What is  the standard normal curve?</h3>

The horizontal axis is approached by the standard normal bend as it extends indefinitely both in directions without ever being touched by it. The center of the bell-shaped, z=0 standard normal curve. Between z=3 and z=3, almost the entire area underneath the standard normal curve is located.

<h3>Use of the standard normal curve:</h3>

Use the normal distribution's standard form to calculate probability. Since the standard normal distribution is indeed a probability distribution, the probability that a variable will take on a range of values is indicated by area of the curve between two points. 100% or 1 is the total area beneath the curve.

<h3>According to the given data:</h3>

the region to the left of the standard normal curve,

z=0.49

To the right of,

z = 2.05

So,

The area will be:

= P[z < 0.49] + P[ z >2.05]

= P[z < 0.49] + 1 -  P[ z < 2.05]

= .6879 + 1 - .9798

= 0.7081

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

To know more about standard normal curve visit:

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I understand that the question you are looking for is:

Find the area under the standard normal curve to the left of z = 0.49 and to the right of z = 2.05. Round your answer to four decimal places, if necessary.

4 0
2 years ago
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